Integral von $$$2 \tan^{2}{\left(\theta \right)}$$$

Bestimme $$$\int 2 \tan^{2}{\left(\theta \right)}\, d\theta$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ mit $$$c=2$$$ und $$$f{\left(\theta \right)} = \tan^{2}{\left(\theta \right)}$$$ an:

$${\color{red}{\int{2 \tan^{2}{\left(\theta \right)} d \theta}}} = {\color{red}{\left(2 \int{\tan^{2}{\left(\theta \right)} d \theta}\right)}}$$

Sei $$$u=\tan{\left(\theta \right)}$$$.

Dann gelten $$$\theta=\operatorname{atan}{\left(u \right)}$$$ und $$$d\theta=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$ (die Schritte sind » zu sehen).

Also,

$$2 {\color{red}{\int{\tan^{2}{\left(\theta \right)} d \theta}}} = 2 {\color{red}{\int{\frac{u^{2}}{u^{2} + 1} d u}}}$$

Forme den Bruch um und zerlege ihn:

$$2 {\color{red}{\int{\frac{u^{2}}{u^{2} + 1} d u}}} = 2 {\color{red}{\int{\left(1 - \frac{1}{u^{2} + 1}\right)d u}}}$$

Gliedweise integrieren:

$$2 {\color{red}{\int{\left(1 - \frac{1}{u^{2} + 1}\right)d u}}} = 2 {\color{red}{\left(\int{1 d u} - \int{\frac{1}{u^{2} + 1} d u}\right)}}$$

Wenden Sie die Konstantenregel $$$\int c\, du = c u$$$ mit $$$c=1$$$ an:

$$- 2 \int{\frac{1}{u^{2} + 1} d u} + 2 {\color{red}{\int{1 d u}}} = - 2 \int{\frac{1}{u^{2} + 1} d u} + 2 {\color{red}{u}}$$

Das Integral von $$$\frac{1}{u^{2} + 1}$$$ ist $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

$$2 u - 2 {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}} = 2 u - 2 {\color{red}{\operatorname{atan}{\left(u \right)}}}$$

Zur Erinnerung: $$$u=\tan{\left(\theta \right)}$$$:

$$- 2 \operatorname{atan}{\left({\color{red}{u}} \right)} + 2 {\color{red}{u}} = - 2 \operatorname{atan}{\left({\color{red}{\tan{\left(\theta \right)}}} \right)} + 2 {\color{red}{\tan{\left(\theta \right)}}}$$

Daher,

$$\int{2 \tan^{2}{\left(\theta \right)} d \theta} = 2 \tan{\left(\theta \right)} - 2 \operatorname{atan}{\left(\tan{\left(\theta \right)} \right)}$$

Vereinfachen:

$$\int{2 \tan^{2}{\left(\theta \right)} d \theta} = 2 \left(- \theta + \tan{\left(\theta \right)}\right)$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{2 \tan^{2}{\left(\theta \right)} d \theta} = 2 \left(- \theta + \tan{\left(\theta \right)}\right)+C$$

$$$\int 2 \tan^{2}{\left(\theta \right)}\, d\theta = 2 \left(- \theta + \tan{\left(\theta \right)}\right) + C$$$A